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Pascal's law, pressure, consequences

  1. Feb 24, 2011 #1
    Yes equations say its the same pressure, but its soooooooo non intuitive!!!!

    Can you please explain, and try to near this problem to me, i can't sleep!!

    [PLAIN]http://img821.imageshack.us/img821/7469/75687957.jpg [Broken]

    How come the pressure at A and B is the same, although we don't have same amount of water ... pascal's law says it is... But i don't get it...
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 24, 2011 #2


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    I assume these are two containers filled with a liquid like water, with the top open to the air.

    To see that the pressure at depth [itex]h[/itex] is always [itex]\rho g h [/itex] (i.e. the weight of a column of water above that depth, up to the free surface), forget about the shape of the container.

    Think about a little cube of water somewhere in the container. This is a statics situation. There is no flow or movement of the water. So the forces on that little cube have to be in equilibrium.

    Therefore, the force on the bottom of the cube must be equal to the force on the top of the cube, plus the weight of the cube itself. It this was not true, the little cube of water would move, because of the unbalanced forces acting on it.

    Write the force on the top and bottom as (pressure x area), and the weight as (g x mass) = (g x density x volume) = (g x density x area x height), cancel out the areas, and you get Pascal's law.
  4. Feb 24, 2011 #3
    That was my assumption too.

    I've removed my post because I think yours covers it far better.
  5. Feb 25, 2011 #4
    [PLAIN]http://img694.imageshack.us/img694/8074/55627195.jpg [Broken]

    What bothers me, is that how come the pressure is the same. Look, i figure that tube of water in B tank is longer, i figure there is more fluid there ergo more pressure at the bottom...

    But at A tank that tube is much smaller in my intuition pressure would be lower that in the tank B...

    How do fluids add up? do they stack each other like, i don't know layers and each layer is affected by the one before, or are you watching only discreet sections of the fluid by it self.

    I think this misunderstanding what i have is mainly because i am used to solids...
    Last edited by a moderator: May 5, 2017
  6. Feb 25, 2011 #5


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    The simplest answer to that, I think, is that, if the pressure were not the same, then water would FLOW to equalise it. Where would it go in your diagram? The bottom of the 'boot' is already full.
    Remember, where you have a fluid, the pressure acts in all directions - sideways as well as up or down. The upwards pressure on the underside of the 'boot' shape is the same as the sideways pressure on the sides at that height and equal to the downwards pressure on the water below.

    ANother way to think of it: if the pressure were different in different directions then you could make a craft that would travel around, utilising this force difference and not use any energy. And we all know that's a big no no.
  7. Feb 25, 2011 #6

    jack action

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    What is pressure?

    Pressure is the force divided by the area perpendicular to its application. The interesting thing about pressure is that, when everything is in equilibrium, it is equal in every direction.

    The force in question here is the weight. For a given area on the ground, the weight of the atmospheric air column acting on this area will give us the atmospheric pressure. So when we say that the atmospheric pressure is 14,7 pounds per squared inch, that means that for every square of 1" X 1", the quantity of air over that surface weights 14,7 lb.

    This is why when we go up in altitude, the pressure drops: the column of air over us is shorter.

    At ground level, two side-by-side areas have the same air pressure acting on them, therefore there is equilibrium, the air does not displace sideways and pressure is equal in all directions.

    Now if we replace some of that air with, say, water, the new weight will be the weight of the (now shorter) air column PLUS the weight of the water column. Just like the case for air only, if there is no displacement, this means that the pressure has to be equal in every directions at the bottom of the water column. This is why the pressure is equal in both of your example: if at one place there is a column of water of equal height (and the same weight of air over it), the pressure at the bottom will be the same in both cases. And because there is equilibrium, in a horizontal plane (same height), the pressure has to be the same everywhere in that plane.

    By the way the proper equation is:

    [tex]P = P_{o} + \rho gh[/tex]

    Where [tex]\rho gh[/tex] is the pressure due to a column of a given height and density (a column of water for example) and [tex]P_{o}[/tex] is the pressure acting over that column (the column of air over that column of water for example).
  8. Feb 25, 2011 #7
    I am beginning to form a slight picture... now new question has risen, would the pressure be the same if there was no air at all? just 2 examples but no air on the outside
  9. Feb 25, 2011 #8

    jack action

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    Yes, they would be the same, although of a smaller magnitude since now [tex]P_{o}[/tex] is equal to zero in both cases.
  10. Feb 25, 2011 #9
    This weight-force hypothesis only holds under conditions of equilibrium. In the free atmosphere, a wind blowing across a surface drops the pressure on that surface even though the mass density of the overlying column is unchanged.

    That is why heavy sheets of plywood get pushed out of the beds of pickup trucks. The ambient pressure under them is greater than the dynamically-reduced pressure over them. Similarly, wind blowing past a vented building will cause an undampened barometer inside to "pump" with variations in the wind velocity.

    Consequently, one may assume that if the wind is blowing at any elevation at all, a barometer at the base of the column of air will not correctly measure the weight of the overlying air.

    I like the kinetic gas explanation of gas pressure better. Pressure is simply the mean number of molecular impacts per unit area times the mean impulse transferred per impact.
  11. Feb 26, 2011 #10
    Yes yes, dynamics happen when pressure is not the same i get that. But what I am asking, how come quantitatively pressure is the same in A and B at those spots, what is bugging me how come the water on top of B doesn't affect pressure at bottom. That's my problem.
  12. Feb 26, 2011 #11

    jack action

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    water pressure.jpg

    At any point (A, B, C or D), and for that point only, the pressure is the same in all directions (red lines).

    At point C:

    [tex]P_{C} = P_{o} + \rho gh_{1}[/tex]

    Where [tex]P_{o}[/tex] is the atmospheric pressure.

    At point B:

    [tex]P_{B} = P_{C} + \rho gh_{2}[/tex]


    [tex]P_{B} = P_{o} + \rho gh_{1} + \rho gh_{2} = P_{o} + \rho g \left( h_{1} + h_{2} \right)[/tex]

    At point D, because it is at the same level than point C, their «left» and «right» pressures must be equal. Therefore:

    [tex]P_{D} = P_{C}[/tex]

    For point A, two methods can be used. Method 1 the same as the one for point D, for the same reason:

    [tex]P_{A} = P_{B}[/tex]

    Method 2 is the following:

    [tex]P_{A} = P_{D} + \rho gh_{2}[/tex]

    But, we already established that [tex]P_{D} = P_{C}[/tex] and [tex]P_{C} = P_{o} + \rho gh_{1}[/tex], So:

    [tex]P_{A} = P_{o} + \rho gh_{1} + \rho gh_{2} = P_{o} + \rho g \left( h_{1} + h_{2} \right)[/tex]


    [tex]P_{A} = P_{B}[/tex]

    The pressure at the bottom is always the weight that is supported PLUS the pressure over that weight. Because, in our case, there is no lateral acceleration, there is no «weight» to be supported ([tex]g = 0[/tex]), so the term [tex]\rho gh = 0[/tex] when we look at it «sideways».
  13. Feb 26, 2011 #12
    I am getting some things mixed up here. When go down the ocean what happens?

    We have low pressure at top and high pressure at bottom, what why doesnt water then rush from bottom to top?

    And another question, WHY is pressure distributed in all directions evenly(without gravity case, for simplicity), Can someone go to atomic scale and explain this?
    Last edited: Feb 26, 2011
  14. Feb 26, 2011 #13

    jack action

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    Because that pressure difference is used to oppose the gravity force, hence no motion.

    If there is equilibrium (no motion), all forces must be equal and opposite in all directions. Imagine a cube of equal side and a force acting on each side. Do the 3-D free body diagram and you'll find out that all forces are equal on each side, hence the pressure is also the same because of the area similarity. If it wasn't, there would be motion or at least the cube would deform.

    The key word in all these problems is equilibrium.
  15. Feb 26, 2011 #14
    I think i got it, thanks all
  16. Apr 1, 2013 #15

    the answer to you question was given by pascal himself
    In the first Diagram:
    The molecules of water are in continuous motion. During their motion they hit the upeer of of the container.
    According to newtons third law "for every action there is an equal and opposite reaction".
    so the upper surface of the container will exert a force on the liquid so the net force on the bottom of container will be the "weight of the liquid plus" the reaction force by the upper surface of container
    so from this we conclude that the pressure at the bottom of both container will be same.

    The is an excellent question..... the question you have asked is related to topic called "hydrostatic paradox". These type of vases in which hydro static paradox occur is called "hydro static vases"

    Hope this if you still don't understand please do let me know.....i will make it more clear for you.
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